world youth mathematics intercity · pdf filemathematics competitions vol 21 no 1 2008 world...

Mathematics Competitions Vol 21 No 1 2008

World Youth Mathematics Intercity

Competition

Simon Chua, Andy Liu & Bin Xiong

Simon Chua is the founder and pres-ident of the Mathematics Trainers’Guild, Philippines. The first Filipinoto win the Paul Erdos National Awardfrom the World Federation of NationalMathematics Competitions for his dis-tinguished and sustained contributionto the enrichment of mathematics edu-cation in the Philippines. His ambitionis to transform all mathematics teach-ers in the Philippines to good mathe-matics educators, so that more Filipinokids would achieve international recog-nition in the field of mathematics.

Andy Liu is a professor of mathemat-ics at the University of Alberta inCanada. His research interests spansdiscrete mathematics, geometry, math-ematics education and mathematicsrecreations. He edits the Problem Cor-ner of the MAA’s magazine Math Hori-zons. He was the Chair of the ProblemCommittee in the 1995 IMO in Canada.His contribution to the 1994 IMO inHong Kong was a main reason for hisbeing awarded a David Hilbert Inter-national Award by the World Federa-tion of National Mathematics Compe-titions. He has trained students in allsix continents.

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Bin Xiong is a professor of mathemat-ics at the East China Normal Univer-sity. His research interest is in prob-lem solving and gifted education. Heis a member of the Chinese Mathe-matical Olympiad committee and theProblem Subcommittee. He had servedas the leader of the national teamat the 2005 International Mathemat-ical Olympiad in Mexico. He hadbeen a trainer at the national campmany times. He is involved in the Na-tional Junior High School Competition,the National High School Mathemat-ics Competition, the Western ChinaMathematical Olympiad and the Girls’Mathematical Olympiad.

The World Youth Mathematics Intercity Competition was founded in1999 by Prof. Hsin Leou of the Kaohsiung National Normal Universityin Taiwan. It was designed as an International Mathematical Olympiadfor junior high school students. Each team consists of a leader, a deputyleader and four contestants. They represent their city rather than theircountry, downplaying politics.

For the first two years, the contest was actually held in Kaohsiung, andthe participating teams were all from South East Asia. The hosts in2001 and 2002 were the Philippines and India, respectively. It was notheld in 2003 during the height of the SARS scare. In 2004, the host wasMacau.

In 2005, the contest returned to Kaohsiung. Much progress has beenmade since the beginning. China had joined the rank, and in fact theChinese city of Wen Zhou hosted the 2006 contest. In 2007, when thecontest was held in Chang Chun, China, there were sixty-five teams,including those from Canada, Iran, South Africa and the United Statesof America.

There is now a permanent WYMIC board. The president is Mr. Wen-Hsien Sun of Chiu Chang Mathematics Foundation in Taipei, and thesecretary is Mr. Simon Chua of the Mathematics Trainers’ Guild in

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Zamboanga. Added to the board each year are the local organizingcommittee and the local problem committee. Prof. Zonghu Qiu ofAcademia Sinica in Beijing is the advisor for the former, and Prof. AndyLiu and Prof. Bin Xiong are advisors for the latter.

The event typically lasts five days in late July. Day 1 is for arrival. Day2 is taken up with registration, the opening ceremony and team leadermeetings. Day 3 is the actual competition, with an individual contest inthe morning and a team contest in the afternoon. Day 4 is set aside forexcursion, with a banquet in the evening. Day 5 is for departure afterthe closing ceremony. Slight variation occurs from year to year.

Gold, silver and bronze medals are awarded for the individual contest,as well as honorable mentions. The teams are drawn into a number ofgroups. Within each group, gold, silver and bronze medals are awardedfor the team contest. A second set of medals are awarded on the basisof the sum of the best three scores in the individual contest. Finally,a team is declared the grand champion, based on the sum of these twoteam scores. The next two teams also received trophies. In addition,there are two non-academic team awards, one for the best behaviourand one for popularity.

A most special feature of the WYMIC is the Cultural Evening. Eachteam must perform on stage for about five minutes, to showcasetheir ethnic identities. The performances vary from instrumental andchoral music to power-point presentations of scenery. This cementsinternational friendship and fosters mutual understanding. Two moreteam awards are handed out, one for the best performance and one forinnovation

In 2009, the contest will be hosted by South Africa in Durban. Itis a golden opportunity for European and other African teams tojoin in. For further information, contact Prof. Gwen Williamsat [email protected]. To conclude this paper, we append theproblems used in the 2007 contest.

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1 Individual Contest

Time limit: 120 minutes 2007/7/23 Changchun, China

Section I

In this section, there are 12 questions, fill in the correct answers in thespaces provided at the end of each question. Each correct answer is worth5 points.

1. Let An be the average of the multiples of n between 1 and 101.Which is the largest among A2, A3, A4, A5 and A6?

Solution. The smallest multiple of n is of course n. Denote by anthe largest multiple of n not exceeding 101. Then An = (n+an)/2.Hence A2 = A3 = 51, A4 = 52, A5 = 52.5 and A6 = 51, and thelargest one is A5.

2. It is a dark and stormy night. Four people must evacuate from anisland to the mainland. The only link is a narrow bridge whichallows passage of two people at a time. Moreover, the bridge mustbe illuminated, and the four people have only one lantern amongthem. After each passage to the mainland, if there are still peopleon the island, someone must bring the lantern back. Crossing thebridge individually, the four people take 2, 4, 8 and 16 minutesrespectively. Crossing the bridge in pairs, the slower speed is used.What is the minimum time for the whole evacuation?

Solution. Exactly five passages are required, three pairs to themainland and two individuals back to the island. Let the fastesttwo people cross first. One of them brings back the lantern.Then the slowest two people cross, and the fastest people on themainland brings back the lantern, The final passage is the same asthe first. The total time is 4+2+16+4+4 = 30 minutes. To showthat this is minimum, note that the three passages in pairs take atleast 16 + 4 + 4 = 24 minutes, and the two passages individuallytake at least 4 + 2 = 6 minutes.

3. In triangle ABC, E is a point on AC and F is a point on AB.BE and CF intersect at D. If the areas of triangles BDF , BCD

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and CDE are 3, 7 and 7 respectively, what is the area of thequadrilateral AEDF?

Solution. Since triangles BCD and CDE have equal areas,BD=DE. Hence the area of triangle DEF is also 3. Let thearea of triangle EFA be x. Then x

6 = AFBF = x+3+7

3+7 . It followsthat 10x = 6x + 60 so that x = 15. The area of the quadrilateralAEDF is 15 + 3 = 18.

4. A regiment had 48 soldiers but only half of them had uniforms.During inspection, they form a 6 × 8 rectangle, and it was justenough to conceal in its interior everyone without a uniform. Later,some new soldiers joined the regiment, but again only half of themhad uniforms. During the next inspection, they used a differentrectangular formation, again just enough to conceal in its interioreveryone without a uniform. How many new soldiers joined theregiment?

Solution. Let the dimensions of the rectangle be x by y, withx � y. Then the number of soldiers on the outside is 2x + 2y − 4while the number of those in the interior is (x − 2)(y − 2). Fromxy − 2x − 2y + 4 = 2x + 2y − 4, we have (x − 4)(y − 4) =xy − 4x − 4y + 16 = 8. If x − 4 = 2 and y − 4 = 4, we obtainthe original 6 × 8 rectangle. If x − 4 = 1 and y − 4 = 8, weobtain the new 5 × 12 rectangle. Thus the number of new soldiersis 5 × 12 − 6 × 8 = 12.

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5. The sum of 2008 consecutive positive integers is a perfect square.What is the minimum value of the largest of these integers?

Solution. Let a be the smallest of these integers. Then

a + (a + 1) + (a + 2) + . . . + (a + 2007) = 251 · (2a + 2007) · 22.

In order for this to be a perfect square, we must have 2a + 2007 =251n2 for some positive integer n. For n = 1 or 2, a is negative.For n = 3, we have a = 126 so that a + 2007 = 2133 is the desiredminimum value.

6. The diagram shows two identical triangular pieces of paper A andB. The side lengths of each triangle are 3, 4 and 5. Each triangle isfolded along a line through a vertex, so that the two sides meetingat this vertex coincide. The regions not covered by the foldedparts have respective areas SA and SB. If SA + SB = 39 squarecentimetres, find the area of the original triangular piece of paper.

Solution. In the first diagram, the ratio of the areas of the shadedtriangle and one of the unshaded triangle is (5 − 3) : 3 so that SA

is one quarter of the area of the whole triangle. In the seconddiagram, the ratio of the areas of the shaded triangle and one ofthe unshaded triangle is (5 − 4) : 4 so that SB is one ninth of thearea of the whole triangle. Now 1

4 + 19 = 13

36 . Hence the area of thewhole triangle is 36

1339 = 108 square centimetres.

7. Find the largest positive integer n such that 31024 − 1 is divisibleby 2n.

Solution. Note that

31024 − 1 = (3512 + 1)(3256 + 1)(3128 + 1) . . . (3 + 1)(3 − 1).

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All 11 factors are even, and 3+1 is a multiple of 4. Clearly 3 − 1is not divisible by 4. We claim that neither is any of the other 9.When the square of an odd number is divided by 4, the remainderis always 1. Adding 1 makes the remainder 2, justifying the claim.Hence the maximum value of n is 12.

8. A farmer has four straight fences, with respective lengths 1, 4, 7and 8 metres. What is the maximum area of the quadrilateral thefarmer can enclose?

Solution. We may assume that the sides of lengths 1 and 8 areadjacent sides of the quadrilateral, as otherwise we can flip overthe shaded triangle in the first diagram. Now the quadrilateralmay be divided into two triangles as shown in the second diagram.In each triangle, two sides have fixed length. Hence its area ismaximum if these two sides are perpendicular to each other. Since12+82 = 42+72, both maxima can be achieved simultaneously. Inthat case, the area of the unshaded triangle is 4 and the area of theshaded triangle is 14. Hence the maximum area of the quadrilateralis 18.

9. In the diagram (on the next page), PA = QB = PC = QC =PD = QD = 1, CE = CF = EF and EA = BF = 2AB.Determine BD.

Solution. Let M be the midpoint of EF . By symmetry, D lieson CM . Let BM = x. Then FM = 5x, CF = 10x, CM = 5

√5x

and BC = 2√

19x. It follows that ABBC = 1√

19. Now Q is the

circumcentre of triangle BCD. Hence ∠BQD = 2∠BCD =∠BCA. Since both QDB and CAB are isosceles triangles, theyare similar to each other. It follows that BD

QB = ABBC = 1√

19, so that

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BD = 1√19

.

10. Each of the numbers 2, 3, 4, 5, 6, 7, 8 and 9 is used once to fill inone of the boxes in the equation below to make it correct. Of thethree fractions being added, what is the value of the largest one?

Solution. We may assume that the second numerator is 5 and thethird 7. If either 5 or 7 appears in a denominator, it can never beneutralized. Since the least common multiple of the two remainingnumbers is 8 × 9 = 72, we use 1

72 as the unit of measurement.Now one of the three fractions must be close to 1. This can onlybe 5

2×3 or 72×4 . In the first case, we are short 12 units. Of this,

7 must come from the third fraction so that 5 must come fromthe first fraction. This is impossible because the first fraction hasnumerator 1 and 5 does not divide 72. In the second case, we areshort 9 units. Of this, 5 must come from the second fraction sothat 4 must come from the third. This can be achieved as shownin the equation below. Hence the largest of the three fractions has

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value 78 .

11. Let x be a positive number. Denote by [x] the integer part of xand by {x} the decimal part of x. Find the sum of all positivenumbers satisfying 5{x} + 0.2[x] = 25.

Solution. The given equation may be rewritten as {x} = 125−[x]25 .

From 0 � {x} < 1, we have 100 < [x] � 125. For each solutionx, x = [x] + {x} = 5 + 24

25 [x]. It follows that the desired sum is5(25) + (24/25)(101 + 102 + 103 + . . . + 125) = 2837.

12. A positive integer n is said to be good if there exists a perfectsquare whose sum of digits in base 10 is equal to n. For instance,13 is good because 72 = 49 and 4 + 9 = 13. How many goodnumbers are among 1, 2, 3, . . . , 2007?

Solution. If a positive integer is a multiple of 3, then its squareis a multiple of 9, and so is the sum of the digits of the square.If a positive integer is not a multiple of 3, then its square is 1more than a multiple of 3, and so is the sum of the digits of thesquare. Now the square of 9 . . . 9 with m 9s is 9 . . . 980 . . .01, withm − 1 9s and 0s. Its digit sum is 9m. Hence all multiples of 9are good, and there are 2007

9 = 223 of them not exceeding 2007.On the other hand, the square of 3 . . . 35 with m 3s is 1212..1225with m sets of 12. Its digit sum is 3m + 7. Since 1 and 4 are alsogood, all numbers 1 more than a multiple are good, and there are2007

3 = 669 of them. Hence there are altogether 223 + 669 = 992good numbers not exceeding 2007.

Section II

Answer the following 3 questions, and show your detailed solution in thespace provided after each question. Each question is worth 20 points.

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1. A 4× 4 table has 18 lines, consisting of the 4 rows, the 4 columns,5 diagonals running from southwest to northeast, and 5 diagonalsrunning from northwest to southeast. A diagonal may have 2, 3 or4 squares. Ten counters are to be placed, one in each of ten of thesixteen cells. Each line which contains an even number of countersscores a point. What is the largest possible score?

The maximum score is 17, as shown in the placement in thediagram below. The only line not scoring a point is marked.

We now prove that a perfect score of 18 points leads to acontradiction. Note that the five diagonals in the same directioncover all but two opposite corner cells. These two cells must eitherbe both vacant or both occupied. Note also that we must have acompletely filled row, and a completely filled column. We considerthree cases.

Case 1. All four corner cells are vacant.We may assume by symmetry that the second row and the secondcolumn are completely filled. Then we must fill the remaining innercells of the first row, the fourth row, the first column and the fourthcolumn. These requires eleven counters.

Case 2. Exactly two opposite corner cells are vacant.By symmetry, we may assume that one of them is on the first rowand first column, and the other is on the fourth row and fourthcolumn. Then we must have exactly one more occupied inner cellon each of the first row, the first column, the fourth row and thefourth column. This means that all four cells in the interior of thetable are filled. By symmetry, we may assume that the completely

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filled row is the second. It is impossible to score both the diagonalsof length 2 which intersect the second row.

Case 3. All four corner cells are occupied.We claim that the completely filled row must be either the firstor the fourth. Suppose to the contrary it is the second. Then wemust fill the first column and the fourth column, thus using up allten counters. Now there are several diagonals which do not yieldscores. This justifies the claim. By symmetry, we may assume thatthe first row and the first column are completely filled. To score allrows and columns, the remaining two counters must be in the fourinterior cells. Again, some of the diagonals will not yield scores.

2. There are ten roads linking all possible pairs of five cities. It isknown that there is at least one crossing of two roads, as illustratedin the diagram below on the left. There are nine roads linking eachof three cities to each of three towns. It is known that there is alsoat least one crossing of two roads, as illustrated in the diagrambelow on the right. Of the fifteen roads linking all possible pairs ofsix cities, what is the minimum number of crossings of two roads?

Solution. The minimum number of crossing of two roads is three,as illustrated in the diagram below.

Suppose at most two crossings of two roads are needed. If we closeone road from each crossing, the remaining ones can be drawnwithout any crossing. We consider two cases.

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Case 1. The two roads closed meet at a city.Consider the other five cities linked pairwise by ten roads, none ofwhich has been closed. It is given that there must be a crossing oftwo roads, which is a contradiction.

Case 2. The two roads closed do not meet at a city.Choose the two cities linked by one of the closed roads, and athird city not served by the other closed road. Call these threecities towns. Each is linked to each of the remaining three citiesby a road. It is given that there must be a crossing of two roads,which is a contradiction.

3. A prime number is called an absolute prime if every permutationof its digits in base 10 is also a prime number. For example: 2,3, 5, 7, 11, 13 (31), 17 (71), 37 (73) 79 (97), 113 (131, 311), 199(919, 991) and 337 (373, 733) are absolute primes. Prove that noabsolute prime contains all of the digits 1, 3, 7 and 9 in base 10.

Solution. Let N be an absolute prime which contains all of thedigits 1, 3, 7 and 9 in base 10. Let L be any number formed fromthe remaining digits. Consider the following seven permutations ofN : 10000L + 7931, 10000L + 1793, 10000L + 9137, 10000L + 7913,10000L + 7193, 10000L + 1973 and 10000L + 7139. They havedifferent remainders when divided by 7. Therefore one of them is

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a multiple of 7, and is not a prime. Hence N is not an absoluteprime.

2 Team Contest2007/7/23 Changchun, China

1. Use each of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once tofill in the nine small circles in the Olympic symbol below, so thatthe numbers inside each large circle is 14.

Solution. The sum of the nine numbers is 1 + 2 + 3 + 4 + 5 +6 + 7 + 8 + 9 = 45. The total sum of the numbers in the five largecircles is 14 · 5 = 70. The difference 70 − 45 = 25 is the sum ofthe four numbers in the middle row, because each appears in twolarge circles. The two numbers at one end must be 9 and 5 whilethe two numbers at the other end must be 8 and 6. Consider thetwo numbers at the end of the middle row. Clearly they cannotbe 5 and 6. If they are 5 and 8, the other two numbers must sumto 12. With 5 and 8 gone, the only possibility is 9 and 3, but 9cannot be in the inner part of the middle row. If they are 9 and 8,the other two numbers must sum to 8. Since neither 5 nor 6 canappear in the inner part of the middle row, the only possibility is7 and 1. However, 7 cannot be in the same large circle with either9 or 8. It follows that the two numbers at the end of the middlerow are 9 and 6, and the other two numbers sum to 10. The onlypossibility is 7 and 3, and 7 must be in the same large circle with 6.

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The remaining numbers can now be filled in easily, and the 9-digitnumber formed is 861743295.

2. The diagram below shows fourteen pieces of paper stacked on topof one another. Beginning on the pieces marked B, move frompiece to adjacent piece in order to finish at the piece marked F .The path must alternately climb up to a piece of paper stackedhigher and come down to a piece of paper stacked lower. Thesame piece may be visited more than once, and it is not necessaryto visit every piece. List the pieces of paper in the order visited.

Solution. We construct below a diagram which is easier to use.An arrow from one piece of paper to another represents comingdown from the first to the second. Note that each of M and N isconnected to 7 other pieces, each of D and J is connected to 4 otherpieces, while each of the others is connected to 3 other pieces. Thepath we seek consists of alternately going along with the arrow andgoing against it. Of the three arrows at A, the one from B cannotbe used as otherwise we would be stuck at A. Equally useless are

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the arrows from L to M , from M to K, from I to N , from N toH , from G to F , from E to F , and from C to D. In the diagrambelow, they are drawn as single arrows while the useable ones aredrawn as double arrows.

From B, if we climb up to C, then we will go down to M , but wecould have gone down to M directly. Once in M , we have a choiceof C or J , but C leads back to B, and we will get stuck there.From J , we have to move onto K, L, A, M , D, E, N , G, H , I, J ,N and F . Thus the path is BMJKLAMDENGHIJNF .

3. There are 14 points of intersection in the seven-pointed star in thediagram (on the next page). Label these points with the numbers1, 2, 3, . . . , 14 such that the sum of the labels of the four points oneach line is the same. Give one of solution.

Solution. Each of the points of intersection lies on exactly twolines. Hence the common sum is given by 2(1+2+3+. . . +14)/7 =30. We claim that the smallest label on the same line with 14 is1 or 2. Otherwise, the sum of the labels of the two lines is atleast 14 + 14 + 3 + 4 + 5 + 6 + 7 + 8 = 61 > 30 + 30, which is acontradiction. Similarly, the smallest label on the same line with13 is 1, 2 or 3. In view of these observations, we put 14 on a linewith 1 and on another line with 2, and 13 on a line with 1 andon another line with 2 or 3. This leads to the three labeling on

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the top row. The labeling on the bottom row are the complementsof the corresponding ones in the top row, that is, each label k isreplaced by 15 − k.

4. Mary found a 3-digit number that, when multiplied by itself,produced a number which ended in her 3-digit number. Whatis the sum of the numbers which have this property?

Solution. Since 1 × 1 = 1, 5 × 5 = 25 and 6 × 6 = 36, the lastdigit of the 3-digit number must be either 1, 5 or 6.

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No 2-digit number with units digit 1 whose square ends with thatnumber.There is only one 2-digit number with units digit 5 whose squareends with that number and that is 25.There is only one 2-digit number with units digit 6 whose squareends with that number and that is 76.There is only one 3-digit number which ends in 25 whose squareends with that number and that is 625.

There is only one 3-digit number which ends in 76 whose squareends with that number and that is 376.

The sum of these two numbers is 625+376=1001.

5. Determine all positive integers m andn such that m2 +1 is a primenumber and 10(m2 + 1) = n2 + 1.Solution. From the given condition, 9(m2 +1) = (n+m)(n−m).Note that m2 + 1 is a prime number not equal to 3. Hence thereare four cases.

a) n − m = 1, n + m = 9(m2 + 1).Subtraction yields 9m2 + 8 = 2m, which is impossible.

b) n − m = 3, n + m = 3(m2 + 1).Subtraction yields 3m2 = 2m, which is impossible.

c) n − m = 9, n + m = m2 + 1.Subtraction yields 2m = m2 − 8, so that m = 4 and n = 13.Note that m2 + 1 = 17 is indeed a prime number.

d) n − m = m2 + 1, n + m = 9.Subtraction yields −2m = m2−8, so that m = 2 and n = 7.Note that m2 + 1 = 5 is indeed a prime number.

In summary, there are two solutions, (m, n) = (2, 7) or (4, 13).

6. Four teams take part in a week-long tournament in which everyteam plays every other team twice, and each team plays one gameper day. The diagram below on the left shows the final scoreboard,part of which has broken off into four pieces, as shown on thediagram below on the right. These pieces are printed only on oneside. A black circle indicates a victory and a white circle indicatesa defeat. Which team wins the tournament?

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Solution. When reconstructing the broken scoreboard, there aretwo positions where the U-shaped piece can be placed. So as toleave room for the 3×2 rectangle. Once it is in place, the positionsfor the remaining pieces are determined. They are placed so thatthere are two black circles and two white circles in each column.There are two possibilities, as shown in the diagrams below, butin either case, the winner of the tournament is Team C.

7. Let A be a 3 by 3 array consisting of the numbers 1, 2, 3, . . . , 9.Compute the sum of the three numbers on the i-th row of A andthe sum of the three numbers on the j-th column of A. The numberat the intersection of the i-th row and the j-th column of a 3 by 3array B is equal to the absolute difference of these two sums. Forexample, b12 = |(a11 + a12 + a13) − (a12 + a22 + a32)|.Is it possible to arrange the numbers in A so that the numbers inB are also 1, 2, 3, . . . , 9?

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a11 a12 a13 b11 b12 b13

a21 a22 a23 b21 b22 b23

a31 a32 a33 b31 b32 b33

A B

Solution. Let C be defined just like B, except that we use theactual difference instead of the absolute difference. Compute thesum of the nine numbers in C. Each number in A appears twice inthis sum, once with a positive sign and once with a negative sign.Hence this sum is 0. It follows that among the nine numbers inC, the number of those which are odd is even. The same is trueof the nine numbers in B, since taking the absolute value does notaffect parity. Thus it is not possible for the nine numbers in B be1, 2, 3, . . . , 9.

c11 c12 c13

c21 c22 c23

c31 c32 c33

C

8. The diagonals AC and BD of a convex quadrilateral areperpendicular to each other. Draw a line that passes through pointM , the midpoint of AB and perpendicular to CD, draw anotherline through point N , the midpoint of AD and perpendicular toCB. Prove that the point of intersection of these two lines lies onthe line AC.

Solution. Let M , K and N be the respective midpoints of AB,AC and AD. Then MN is parallel to BD, MK is parallel to BCand NK is parallel to CD. Hence AC and the two lines in questionare the altitudes of triangle MNK, and are therefore concurrent.

9. The positive integers from 1 to n (where n > 1 are arranged in a

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line such that the sum of any two adjacent numbers is a square.What is the minimum value of n?

Solution. The minimum value of n is 15. Since n > 1, we mustinclude 2, so that n � 7 because 2+7 = 9. For n = 7, we have threeseparate lines (1, 3, 6), (2, 7) and (4, 5). Adding 8 only lengthen thefirst to (8, 1, 3, 6). Adding 9 now only lengths the second to (2, 7, 9).Hence n � 10. Now 8, 9 and 10 all have to be at the end if we havea single line, because we can only have 8 + 1 = 9, 9 + 7 = 16 and10 + 6 = 16. The next options are 8 + 17 = 25, 9 + 16 = 25 and10+15 = 25. Hence n � 15. For n = 15, we have the arrangement8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7 and 9.

10. Use one of the five colours (R represent red, Y represent yellow, Brepresent blue, G represent green and W represent white) to painteach square of an 8×8 chessboard, as shown in the diagram below.Then paint the rest of the squares so that all the squares of thesame colour are connected to one another edge to edge. What isthe largest number of squares of the same colour as compare to theother colours?

Solution. While it may tempting to colour the entire fourth rowgreen, this will divide the red squares, the yellow squares and theblue squares into two disconnected parts. Obviously, the northeastcorner is to be used to allow the green path to get around the yellowpath. Similarly, the southwest corner is to be used to allow the bluepath to get around the white path, and the southeast corner is to

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be used to allow the yellow path to get around the white path. Infact, we can complete the entire yellow path. Also, the white pathmay as well make full use of the seventh row. This brings us tothe configuration as shown in the diagram below on the left. It isnow not hard to complete the entire configuration, which is shownin the diagram below on the right. The longest path is the greenone, and the number of green squares is 24.

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Simon ChuaMathematics Trainers’ GuildPHILIPPINES

Andy LiuUniversity of AlbertaCANADA

Bin XiongEast China Normal UniversityCHINA

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